Noetherian induction; closed sets that are not finite union of irreducibles

Question

I am following an online course on algebraic geometry. On today's lesson, the lecturer stated the following theorem: Any Noetherian space is a finite union of irreducible subspaces. For the proof, he takes C to be the minimal element of the set of those closed subsets that cannot be represented as a finite union of irreducibles. He then automatically assumes that C has to be reducible, because he says that C irreducible automatically leads to a contradiction. I don't see why this happens. I have proved that C cannot be irreducible (and also that it cannot be irreducible) and therefore the theorem is proved, but I don't see why this happens immediately; it took me a few lines to do it.

Answer 1

$C$ was taken as the minimal element from a certain set. Think about what defines the elements of that set, and why not just $C$, but all elements in that set must be reducible.